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example of quantum commutator algebra
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(Example)
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Here we illustrate a simple example of quantum commutator algebra using a one-dimensional quantum system. Let  be a function of  . The three commutators of  and of each of the functions  ,  , and  may all be identified (to within the factor  ) with the derivative with respect to  of these functions, but they are not the same operators. Indeed, by repeated application of the commutator algebra rule
![$\displaystyle [q_i,G(p_1,\dots,p_R)] = i\hbar \frac{\partial G}{\partial p_i}$](http://images.physicslibrary.org/cache/objects/835/l2h/img9.png "$\displaystyle [q_i,G(p_1,\dots,p_R)] = i\hbar \frac{\partial G}{\partial p_i}$") |
(1) |
we get
In the same way
[1] Messiah, Albert. "Quantum mechanics: volume I." Amsterdam, North-Holland Pub. Co.; New York, Interscience Publishers, 1961-62.
This entry is a derivative of the Public domain work [1].
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"example of quantum commutator algebra" is owned by bloftin.
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This object's parent.
Cross-references: work, domain, volume, quantum mechanics, commutator algebra, operators, commutators, function, system
This is version 2 of example of quantum commutator algebra, born on 2010-02-14, modified 2010-02-14.
Object id is 835, canonical name is ExampleOfQuantumCommutatorAlgebra.
Accessed 497 times total.
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Pending Errata and Addenda
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![$\displaystyle [q,p^2 f(q)] = 2 i \hbar p f(q)$](http://images.physicslibrary.org/cache/objects/835/l2h/img10.png "$\displaystyle [q,p^2 f(q)] = 2 i \hbar p f(q)$")
![$\displaystyle [q,pfp] = i \hbar(fp+pf)$](http://images.physicslibrary.org/cache/objects/835/l2h/img11.png "$\displaystyle [q,pfp] = i \hbar(fp+pf)$")
![$\displaystyle [q,fp^2] = 2 i \hbar f p$](http://images.physicslibrary.org/cache/objects/835/l2h/img12.png "$\displaystyle [q,fp^2] = 2 i \hbar f p$")
![$\displaystyle [p,p^2f] = \frac{\hbar}{i} p^2 f^{\prime}$](http://images.physicslibrary.org/cache/objects/835/l2h/img13.png "$\displaystyle [p,p^2f] = \frac{\hbar}{i} p^2 f^{\prime}$")
![$\displaystyle [p,pfp] = \frac{\hbar}{i} pf^{\prime}p$](http://images.physicslibrary.org/cache/objects/835/l2h/img14.png "$\displaystyle [p,pfp] = \frac{\hbar}{i} pf^{\prime}p$")
![$\displaystyle [p,fp^2] = \frac{\hbar}{i} f^{\prime}p^2$](http://images.physicslibrary.org/cache/objects/835/l2h/img15.png "$\displaystyle [p,fp^2] = \frac{\hbar}{i} f^{\prime}p^2$")